text: The short answer is that it can't.
  A shielded transmitter, connected to an ideal piece of coax, does not generate common mode currents.
  
  The inner and the outer of the cable look connected (and for Direct Current they are), but high frequency currents really cannot pass through the thick metal, they are confined to the inside or the outside.
  Real coax can leak, but in practice it's very little.
  
  Here are some examples.
  
  First - a simple coaxial cable connecting a transmitter to a dummy load. See how the Inside and the Outside of the coax never meet - they really are separate worlds. No RF current can reach the outside of the coax. There are no common-mode currents.
  
  
  
  
  
  
  
  simulate this circuit &ndash; Schematic created using CircuitLab
  
  Second, a shielded transmitter feeding a length of coax, connected to a balanced antenna without a balun, which creates the unbalanced currents.
  
  
  
  
  
  
  
  simulate this circuit
  
  Balanced lines
  
  I should mention that it is not only coax that can be used to generate balanced currents, it's just the example you asked about.
  If the transmitter drives a balanced line with equal and opposite voltages, say with a transformer, there will be no unbalanced voltage, and no common-mode current. Ethernet cables are like this (CAT5 etc, before they were shielded). There is very little radiation from the pair, even though it is carrying significant RF power, because the transformer that feeds it has a high common-mode impedance, and the other end is an identical circuit, not an antenna.
  
  Generating common-mode currents on purpose
  
  It is possible to generate common mode currents at the transmitter, but generally we take pains to avoid doing that. Here is an example where it is desirable - I have a bluetooth-receiver-FM-transmitter gadget in my car, which works best if there's a USB cable plugged in. It uses the USB cable in common mode as the antenna. Separately, inside the shielding, the USB D+ and D-, and 5V and 0V, are available as differential signals to the phone. The same is done in phones that have FM radio receivers, using the headphone cords.
  
  
  
  
  
  
  
  simulate this circuit